Q: The integrating factor of x d y d x -y=x^4-3 x is

aligned & (c) Since x d y d x -y=x^4-3 x \\ & d y d x - y x =x^3-3 aligned Hence I F=e^ P d x =e^ - 1 x d x =e^ - x = 1 x

Q: ( d y d x ) x=y ^2 x+ x, find general solution

Given that aligned & ( d y d x ) x=y ^2 x+ x \\ & x ( d y d x )-y ^2 x= x \\ & d y d x + ( - ^2 x x ) y= x x \\ & Here, P=- ^2 x x and Q= x \\ & IF =e^ P d x =e^ - ^ 2 x x d x \\ & IF =e^ - ( x) =e^ x = x \\ & Now, y IF = (Q IF ) d x+C \\ & y x= x x+C \\ & y x= 1- ^2 x x +C \\ & y x= | cosec x- x|+ x+C aligned

Question 1

The solution of the differential equation (x+1) d y d x -y=e^ 3 x (x+1)^2 is

Accepted Answer

Given, d y d x - y x+1 =e^ 3 x (x+1) Above is linear differential equation of form aligned & d y d x +P x=Q \ & IF =e^ P d x =e^ - 1 x+1 d x =e^ - (1+x) = 1 1+x aligned Solution will be aligned y IF & = Q IF d x \ y 1 1+x & = e^ 3 x (x+1) 1 (1+x) d x \ y 1+x & = e^ 3 x 3 +C^ \ 3 y 1+x & =e^ 3 x +C^ [ C=3 C^ ] aligned

Question 2

If d y d x -y _e 2=2^ x ( x-1) _e 2, then y

Accepted Answer

d y d x -y _e 2=2^ x ( x-1) _e 2 This is linear differential equation I.F. =e^ - _e 2 d x =e^ -x _e 2 =2^ -x then general Solution is y 2^ -x = 2^ -x 2^ x ( x-1) _e 2 d x+c Now let x-x=t ( x-1) d x=d t aligned & y 2^ -x = _e 2 2^t d t+c \ & y 2^ -x =2^t+c \ & y=2^ x+t +c 2^x \ & y=2^ x +c 2^x aligned

Question 3

The integrating factor of x d y d x -y=x^4-3 x is

Accepted Answer

aligned & (c) Since x d y d x -y=x^4-3 x \ & d y d x - y x =x^3-3 aligned Hence I F=e^ P d x =e^ - 1 x d x =e^ - x = 1 x

Question 4

The number of solutions of d y d x = y+1 x-1 , when y(1)=2 is

Accepted Answer

Since, d y d x = y+1 x-1 d y y+1 = d x x-1 After integrating on both sides, we have (y+1)= (x-1)- C C(y+1)=(x-1) C= x-1 y+1 If x=1, then y=2, so C=0 Therefore, x-1=0 Hence, there is only one solution.

Question 5

( d y d x ) x=y ^2 x+ x, find general solution

Accepted Answer

Given that aligned & ( d y d x ) x=y ^2 x+ x \ & x ( d y d x )-y ^2 x= x \ & d y d x + ( - ^2 x x ) y= x x \ & Here, P=- ^2 x x and Q= x \ & IF =e^ P d x =e^ - ^ 2 x x d x \ & IF =e^ - ( x) =e^ x = x \ & Now, y IF = (Q IF ) d x+C \ & y x= x x+C \ & y x= 1- ^2 x x +C \ & y x= | cosec x- x|+ x+C aligned

Question 6

If the slope of the tangent to the curve at any point P (x, y) is y x - ^ 2 y x , then the equation of a curve passing through (1, 4 ) is

Accepted Answer

According to the condition, dy dx = y x - ^ 2 y x (i) This is a homogeneous differential equation Substituting y=v x, we get array l v+x dv dx =v- ^ 2 v \ x dv dx =- ^ 2 v \ ^ 2 vdv =- dx x \ v=- x+ C \ y x + x= C array Substituting x =1, y= 4 , we get C =1 Thus, we get ( y x )+ x=1 which is the required solution.

Question 7

The general solution of the differential equation ( ^ -1 y-x ) d y= (1+y^ 2 ) d x is

Accepted Answer

The given differential equation can be written as d x d y + x 1+y^ 2 = ^ -1 y 1+y^ 2 (i) Now, eq. (i) is a linear differential equation of the form d x d y + P _ 1 x= Q _ 1 where P _ 1 = 1 1+y^ 2 and Q _ 1 = ^ -1 y 1+y^ 2 Therefore, I.F =e^ 1 1+y^ 2 d y =e^ ^ -1 y Thus, the solution of the given differential equation is given by x e^ ^ -1 y = ( ^ -1 y 1+y^ 2 ) e^ ^ -1 y dy + C (ii) Let I = ( ^ -1 y 1+y^ 2 ) e^ ^ -1 y dy On substituting ^ -1 y=t, so that ( 1 1+y^ 2 ) dy = dt , we get I= t e^ t d

Question 8

The solution of the differential equation (x+1) d y d x -y=e^ 3 x (x+1)^ 2 is

Accepted Answer

The given equation is d y d x - y x+1 =e^ 3 x (x+1) I.F. =e^ - 1 x+1 d x =e^ - (x+1) = 1 x+1 The solution is y ( 1 x+1 )= e^ 3 x (x+1) 1 x+1 d x+a y x+1 = e^ 3 x d x+a= e^ 3 x 3 +a 3 y x+1 =e^ 3 x +c, c=3 a

Question 9

The solutions of (x+y+1) d y=d x are

Accepted Answer

Putting x+y+1=u, we have d u=d x+d y and the given equation reduces to u(d u-d x)= d x u d u u+1 =d x u- (u+1)=x (x+y+2)=y+ constant x+y+2=C e^ y

Differential Equations — Previous Year Questions & Solutions